Product of Algebraic Expressions – Maharashtra Board Class 7 Solutions for Mathematics (English Medium)
MathematicsGeneral ScienceMaharashtra Board Solutions
Exercise 23:
Solution 1:
Solution 2:
- 2(3 + 5) = 2 × 3 + 2 × 5
- 6(7 + 4) = 6 × 7 + 6 × 4
- 10 (9 – 7) = 10 × 9 – 10 × 7
Solution 3:
The pairs of like terms are:
7y, 2y
5ab, 9ab
11ax2, 10x2a
Solution 4:
Solution 5:
Exercise 24:
Solution 1:
Exercise 25:
Solution 1:
Exercise 26:
Solution 1(1):
Solution 1(2):
Solution 1(3):
Solution 1(4):
Solution 1(5):
Exercise 27:
Solution 1(1):
Solution 1(2):
Solution 1(3):
Solution 1(4):
Solution 1(5):
Solution 1(6):
Solution 1(7):
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Solution 1(8):
Solution 1(9):
Solution 1(10):
Solution 1(11):
Solution 1(12):
Exercise 28:
Solution 1(1):
4(y2 – 2y + 7)
= (4 × y2) – (4 × 2y) + (4 × 7)
= 4y2 – 8y + 28
Solution 1(2):
6(2a2 – 3a – 5)
= (6 × 2a2) – (6 × 3a) – (6 × 5)
= 12a2 – 18a – 30
Solution 1(3):
7x(3x2 – 5x + 8)
= (7x × 3x2) – (7x × 5x) + (7x × 8)
= 21x3 – 35x2 + 56x
Solution 1(4):
(m2 – 4m + 6)2m
= (m2 × 2m) – (4m × 2m) + (6 × 2m)
= 2m3 – 8m2 + 12m
Solution 1(5):
(4p2 + p + 3)9p
= (4p2 × 9p) + (p × 9p) + (3 × 9p)
= 36p3 + 9p2 + 27p
Solution 1(6):
3a(a2 + 4a – 6)
= (3a × a2) + (3a × 4a) – (3a × 6)
= 3a3 + 12a2 – 18a
Solution 1(7):
5q(2a + 3b – 4c)
= (5q × 2a) + (5q × 3b) – (5q × 4c)
= 10aq + 15bq – 20cq
Solution 1(8):
4m(3mn – 2n2 + 5)
= (4m × 3mn) – (4m × 2n2) + (4m × 5)
= 12m2n – 8mn2 + 20m
Solution 1(9):
(x2y2 + xy + 1)xy
= (x2y2 × xy) + (xy × xy) + (1× xy)
= x3y3 + x2y2 + xy
Solution 1(10):
11p2q(p2 – 3pq + 8q2)
= (11p2q × p2) – (11p2q × 3pq) + (11p2q × 8q2)
= 11p4q – 33p3q2 + 88p2q3
Exercise 29:
Solution 1(1):
(3y – 2)(y2 – 5y + 7)
= 3y × (y2 – 5y + 7) – 2 × (y2 – 5y + 7)
= (3y × y2 – 3y × 5y + 3y × 7) – (2 × y2 – 2 × 5y + 2 × 7)
= 3y3 – 15y2 + 21y – 2y2 + 10y – 14
= 3y3 – 17y2 + 31y – 14
Solution 1(2):
(a2 – ab + b2)(a + b)
= (a2 – ab + b2) × a + (a2 – ab + b2) × b
= a3 – a2b + ab2 + a2b – ab2 + b3
= a3 + b3
Solution 1(3):
(2p – 5)(3p2 + p – 2)
= 2p × (3p2 + p – 2) – 5 × (3p2 + p – 2)
= 2p × 3p2 + 2p × p – 2p × 2 – 5 × 3p2 – 5 × p + 5 × 2
= 6p3 + 2p2 – 4p – 15p2 – 5p + 10
= 6p3 – 13p2 – 9p + 10
Solution 1(4):
(m2 – 3mn + n2)(5mn + 4)
= (m2 – 3mn + n2) × 5mn + (m2 – 3mn + n2) × 4
= m2 × 5mn – 3mn × 5mn + n2 × 5mn + m2 × 4 – 3mn × 4 + n2 × 4
= 5m3n – 15m2n2 + 5mn3 + 4m2 – 12mn + 4n2
Solution 1(5):
(9y – 4x)(8y2 – 5xy + 3x2)
= 9y × (8y2 – 5xy + 3x2) – 4x × (8y2 – 5xy + 3x2)
= 9y × 8y2 – 9y × 5xy + 9y × 3x2 – 4x × 8y2 + 4x × 5xy – 4x × 3x2
= 72y3 – 45xy2 + 27x2y – 32xy2 + 20x2y – 12x3
= 72y3 – 77xy2 + 47x2y – 12x3
Solution 1(6):
(4m2 + 6mn + 9n2)(2m – 3n)
= (4m2 × 2m + 6mn × 2m + 9n2 × 2m) – (4m2 × 3n + 6mn × 3n + 9n2 × 3n)
= 8m3 + 12m2n + 18mn2 – 12m2n – 18mn2 – 27n3
= 8m3 – 27n3
Solution 1(7):
(p2 + 2q)(7p2 – pq – 11)
= p2 × (7p2 – pq – 11) + 2q × (7p2 – pq – 11)
= (p2 ×7p2 – p2 × pq – p2 × 11) + (2q × 7p2 – 2q × pq – 2q × 11)
= 7p4 – p3q – 11p2 + 14p2q – 2pq2 – 22q
Solution 1(8):
(ab + a2b + ab2)(a + b)
= (ab + a2b + ab2) × a + (ab + a2b + ab2) × b
= (ab × a + a2b × a + ab2 × a) + (ab × b + a2b × b + ab2× b)
= a2b + a3b + a2b2 + ab2 + a2b2 + ab3
= a2b + a3b + 2a2b2 + ab2 + ab3
Solution 1(9):
(x + 2y)(x2 – 2xy + 4y2)
= x × (x2 – 2xy + 4y2) + 2y × (x2 – 2xy + 4y2)
= x × x2 – x × 2xy + x × 4y2 + 2y × x2 – 2y × 2xy + 2y × 4y2
= x3 – 2x2y + 4xy2 + 2x2y – 4xy2 + 8y3
= x3 + 8y3
Solution 1(10):
(3a – 2b)(9a2 + 6ab + 4b2)
= 3a × (9a2 + 6ab + 4b2) – 2b × (9a2 + 6ab + 4b2)
= 3a × 9a2 + 3a × 6ab + 3a × 4b2 – 2b × 9a2 – 2b × 6ab – 2b × 4b2)
= 27a3 + 18a2b + 12ab2 – 18a2b – 12ab2 – 8b3
= 27a3 – 8b3